Textbook Question
A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. c. What is the pH of the buffer after the addition of 0.020 mol of HNO3?
18. Aqueous Equilibrium
Henderson-Hasselbalch Equation
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A 1.0 L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10⁻⁵. Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl to the original buffer using the Henderson-Hasselbalch equation.
A
pH = 4.56
B
pH = 5.00
C
pH = 4.74
D
pH = 4.92
Verified step by step guidance1
Identify the components of the buffer solution: acetic acid (HC2H3O2) and its conjugate base, sodium acetate (NaC2H3O2). The buffer solution initially contains 0.100 mol of each component in 1.0 L.
Understand the effect of adding HCl to the buffer. HCl is a strong acid that will react with the conjugate base (NaC2H3O2) in the buffer, converting some of it to acetic acid (HC2H3O2).
Calculate the moles of HCl added: 10.0 mL of 1.00 M HCl is equivalent to 0.010 mol of HCl. This will decrease the moles of NaC2H3O2 by 0.010 mol and increase the moles of HC2H3O2 by 0.010 mol.
Use the Henderson-Hasselbalch equation to find the new pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \( \text{pK}_a = -\log(\text{K}_a) \). Calculate \( \text{pK}_a \) using \( \text{K}_a = 1.8 \times 10^{-5} \).
Substitute the new concentrations into the Henderson-Hasselbalch equation: After the addition of HCl, the concentration of NaC2H3O2 becomes 0.090 mol/L and the concentration of HC2H3O2 becomes 0.110 mol/L. Use these values to calculate the pH.
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